Q: A particle moves with simple harmonic motion is a straight line. In first τ sec, after starting from rest it travels a distance ‘ a ‘ and in next τ sec, it travels 2a , in same direction, then

(a)Amplitude of motion is 3a

(b)Time period of oscillations is 8π

(c)Amplitude of motion is 4a

(d)Time period of oscillations is 6π

Ans: (d)

Sol: As particle starting from rest .

x = A cosωt

At t= 0 ; x = A

When t = τ ; x = A – a

When t = 2τ ; x = A – 3a

Therefore ,

A – a = A cosωτ

A – 3a = A cos2ωτ

using formula ;

cos2ωτ = 2cos^{2}ωτ – 1

$\large \frac{A-3a}{A} = 2 (\frac{A-a}{A})^2 – 1$

on solving this we get

A = 2a

2a – a = 2a cosωτ

cosωτ = 1/2

ωτ = π/3

$\large \frac{2\pi}{T}\tau = \frac{\pi}{3}$